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1、页眉内容8.7Chapter 88.1In forward biasThenor(a)1 f1For 10, thenI f 2orVi V2 59.6 mV 60mV(b)1 f 1For 100 , thenI f 2orVi V2 119.3 mV 120 mV2.26 1014cm 38.4一34.5 10 cm4.5 104cm(i) pn Xnor Vap noVa exp 一VtVt lnpn Xnpno0.599 V(ii) n-region - lower doped side8.2_43.214 10 cmn po2 ni10 21.5 10Na15102.8125一 43
2、10 cm337.5 10 cmpno2 ni1.5Nd101015-2 101.125510 cm(i) VaVt ln0.1 Nan po0.6165V(a)Va0.45 V,(ii) p-region - lower doped sideornp Xp(b)Va0.55 V,(c)Va0.55 V8.3n po2niNa1.8 106 24 1016pno2ni1.8 106Nd1610(a)Va0.90 V,10.010310 cm3.951210 cm(b) Va1.10V_ 119.88 103cm1.888.11410 cm4.69 1013cm 310 5 cm 343.24
3、10 cm114.0 10 cm9.03 1014 cm8.521.849A/cmIn AJn Xp10 3 1.849 Aor I n 1.85 mA 一，24.521 A/cm一_ 一3Ip AJpXn 104.521 Ap por I p 4.52 mAp(a) I In Ip 1.85 4.52 6.37 mA8.6For an n p silicon diode or15I S 1.8 10 A(a) For Va0.5 V,orI d 4.36 10 7A(b) For Va0.5 V,or15I D I S 1.8 10 A页眉内容8.8Js 1.568 10 4A/cm 2_4
4、2.44 10or I 0.244 mA1.568 10Js 5.14510(i) I(ii) I(iii) I1.02910112A/cm14A_ 141.029 103.61101.029 101.721.029 108.16exp14exp510 A14exp410 A0.450.02590.550.02590.650.02598.9We haveor we can write this asso thatIn reverse bias, I is negative, so at0.90, we haveorV 59.6 mV8.10Case 1:IsexpVaVtIs 6.305101
5、5126.305 10 mA3.153 10 8 mA/cm 2Case 2:Va s exp 一Vtor1.093 mA922 10 mA/cmCase 3:AJsVa exp VtSoVt lnAJsVa 0.6502 VThen4I s AJs 108.11Case 4: INd12.73Na710(a) From part (a),Nd0.354Naii10Va exp VtIsmA1.200.72exp 0.02591.014 10 12 mA525.07 10 5cm2Na0.07857orNd电 2.828orNd8.12The cross-sectional area is3A
6、 I 10 1042A - 5 10 cmJWe have which yields20_ _ 10 _ ,2Js 2.522 10 A/cmWe can writeWe want or37.071 103 Na7.071 10 a 4.472Ndwhich yields Now We findNd 7.091410andNa 1.0110168.13Plot8.14(a)We have0.101033cm3cm页眉内容0.623 VDpDn-and n 2.4noNowpo0.158.4896 10 Asoor(b) Using Einsteins relation, we can writ
7、e We haveThen59.710 10 A8.17n e nNd and p e pNa(a) The excess hole concentration is givenAlsoThenbyWe find8.15(a) p-side;orEFi Ef 0.329 eVp no2niNdand8.1610 21.5 1016102.25104 3cm2.828 10 4cm2.828 mAlso on the n-side; orEf EFi 0.407 eV(b) We can findDnDpNoworJ SThenorIsWe findorThenorPn3.81 1014exp1
8、250 0.02593204.4264.4260.0259ii1010 156I 1.07 10 A(c) The hole current is orI p 3.278 10 16ThenI sp191.6 105I snVa0.8Vbi232.4 cm /s2 ,8.29 cm /sx42.828 10cm(b) We haveAt x 310 4 cm,2A/cmorJp 30.5966 A/cm1.07 AVd exp Vt(A)_ 141.342 10 A410257V2 10 4.025 10 15 A0.746826V0.8 0.74682610 21.5 10165 100.5
9、9746V1431.56 10 cm54.1981 10 A41.3997 10 A1.820 10 4A(c) We haveWe can determine that3n po 4.5 10 cmLn 10.72 mThen orJno 0.2615A/cmWe can also findand.一-,2J po 1.724 A/cm pThen at x 3 m,or2J n 31.39A/cm8.18(a) Problem 8.7 ornpVaVt ln n po(b) Problem 8.8or Va VtVtln0.1 Naln 2n2 N0.205 VP noln0.1 Ndn：
10、 Nd页眉内容8.19byThe excess electron concentration is givenThe total number of excess electrons isWe may note thatThenWe find that 2 .Dn25 cm /s and Ln 50.0 mAlson po2niNa10 21.5 10158 102.811043cmThen orThen, we find the total number of excess electrons in the p-region to be:一一一_ 一 4(a)Va0.3 V,Np1.5110
11、一_ _5(b)Va0.4 V,Np7.1710Va0.5 V,Np3.401071 kT1(b) Taking the ratioFor Ti300 K,38.61For T2400K,kT1kT20.0259,0.03453,1 kT228.96(i) Germanium:Eg0.66 eVI S2or 1383I S1(ii) Silicon: Eg1.12eVI S2or 1.17105I S18.22PlotSimilarly, the total number of excess holesinLpp nothe n-region is found to be We find that2D p 10.0 cm /s and p10.0 mAlso210 2ni1.5 102.25104cm 3Nd1610ThenSo(a)Va0.3 V,Pn2.41103(b)Va0.4 V,Pn1.15510(c)Va0.5 V,Pn5.451068.23First case: or
